- #1
AStaunton
- 100
- 1
Hi there
Problem is:
Assuming spherical Symmetry, estimate the rate of mass loss from the sun, if at the Earth, the measured velocity of the solar wind is 400km/s and the proton density of the wind is roughly 7 particles/cm^3. Give your answer in solar Masses per year.
My attempted solution:
as proton density is 7particles per cm^3 this equals 7*10^6 particles per m^3.
as the speed is 400km per sec this is 4*10^5m per sec
multiply these values to get proton flux:
(7*10^6)*(4*10^5)=2.8*10^12protons m^-2 s^-1
find the surface area of the relevant sphere, this will have radius of 1AU as this flux is measured at the Earth:
A=4*3.14*(1.49598*10^11)^2=2.810873*10^23m^2
now we can find the total protons passing through the entire surface and of course this equals total protons leaving the sun:
(2.810873*10^23)*(2.8*10^12)=7.87044*10^35 protons s^-1
convert this to kg per second:
say proton mass= 1.6727*10^-27kg
so (7.87044*10^35)*(1.627*10^-27)=1316409794kg s^-1
convert to kg per year:
31536000 seconds per year so:
1316409794*31536000=4.151429928*10^16kg per year
find what fraction this mass is of the mass of sun:
(4.151429928*10^16)/(2*10^30)=2.0757*10^-14
so that means mass lost per year this way is (2.0757*10^-14)solarmasses per year.
**************************
When I first read this question, I thought it would involve some calculations regarding the Eddington Luminosity as this is what governs how much mass is blown away...however, the final method that I used did not need that, so can someone please confirm that I did not miss anything or over simplify?
Also, the question speaks of protons in the solar wind, is it safe to assume (as I did in the solution) that these are the only particles blown away?
Any feedback appreciated.
Problem is:
Assuming spherical Symmetry, estimate the rate of mass loss from the sun, if at the Earth, the measured velocity of the solar wind is 400km/s and the proton density of the wind is roughly 7 particles/cm^3. Give your answer in solar Masses per year.
My attempted solution:
as proton density is 7particles per cm^3 this equals 7*10^6 particles per m^3.
as the speed is 400km per sec this is 4*10^5m per sec
multiply these values to get proton flux:
(7*10^6)*(4*10^5)=2.8*10^12protons m^-2 s^-1
find the surface area of the relevant sphere, this will have radius of 1AU as this flux is measured at the Earth:
A=4*3.14*(1.49598*10^11)^2=2.810873*10^23m^2
now we can find the total protons passing through the entire surface and of course this equals total protons leaving the sun:
(2.810873*10^23)*(2.8*10^12)=7.87044*10^35 protons s^-1
convert this to kg per second:
say proton mass= 1.6727*10^-27kg
so (7.87044*10^35)*(1.627*10^-27)=1316409794kg s^-1
convert to kg per year:
31536000 seconds per year so:
1316409794*31536000=4.151429928*10^16kg per year
find what fraction this mass is of the mass of sun:
(4.151429928*10^16)/(2*10^30)=2.0757*10^-14
so that means mass lost per year this way is (2.0757*10^-14)solarmasses per year.
**************************
When I first read this question, I thought it would involve some calculations regarding the Eddington Luminosity as this is what governs how much mass is blown away...however, the final method that I used did not need that, so can someone please confirm that I did not miss anything or over simplify?
Also, the question speaks of protons in the solar wind, is it safe to assume (as I did in the solution) that these are the only particles blown away?
Any feedback appreciated.